Say I have a 5x5 matrix A and I know it is nilpotent, with nilpotence degree = 3.
I compute its null-spaces for $A$, $A^2$,and $A^3$, call each null-space K1, K2, K3, respectively.
I find that K1 (i.e., the eigenspace of A), is of dimension =2. So, I have two independent eigenvectors that span K1.
Similarly, for K2, I find it is of dimension 3, so that dim(K2\K1) = 1.
Finally, dim(K3) = 5, so that dim(K3 \ K2) = 2.
Here is my confusion: I have two linearly independent vectors that span the space K3\K2. With these two vectors in the "top layer", I can then carve out two cyclic subspaces of dimension 3, since each vector gets killed only when I apply A to it three times.
But this gives me a factorization of V into two cyclic subspaces, with total dimension = ... 6? But dim(V) =5.
What have I assumed incorrectly?
Thanks,
If $d_k$ is the nullity of $A^k$ then the sequence $$d_0=0,\,d_1,\,d_2,\,d_3,\,\ldots$$ has two important properties:
That is, the nullities must be (non-strictly) increasing, and the successive differences must be (non-strictly) decreasing.
Your nullity sequence $0,2,3,5$ violates the second of these and hence is not possible.
Sketch of proof for the second property. Take a basis for $\ker(A^k)$, extend to a basis for $\ker(A^{k+1})$ and then to $\ker(A^{k+2})$, so we get say $$\def\u{{\bf u}}\def\v{{\bf v}}\def\w{{\bf w}} \{\u_1,\ldots,\u_p,\v_1,\ldots,\v_q,\w_1\ldots,\w_r\}\ .$$ We need to prove that $r\le q$. To do this consider the vectors $\w_1,\ldots,\w_r$ and write $$\def\x{{\bf x}}\def\y{{\bf y}} A\w_j=\x_j+\y_j\quad\hbox{with}\quad \x_j\in span\{\u_1,\ldots,\u_p\},\ \y_j\in span\{\v_1,\ldots,\v_q\}\ .$$ Then you can show that the $\y_j$ are linearly independent, so we have $r$ independent vectors in a space spanned by $q$ vectors, so $r\le q$ as required.